题目
5、常数项级数sum_(n=1)^infty(sqrt(n+2)-2sqrt(n+1)+sqrt(n))=____.
5、常数项级数$\sum_{n=1}^{\infty}(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n})$=____.
题目解答
答案
将通项重写为：
\[ a_n = (\sqrt{n+2} - \sqrt{n+1}) - (\sqrt{n+1} - \sqrt{n}) \]
求前 $N$ 项和 $S_N$：
\[ S_N = \sum_{n=1}^{N} a_n = (\sqrt{N+2} - \sqrt{N+1}) - (\sqrt{2} - 1) \]
取极限：
\[ \lim_{N \to \infty} S_N = \lim_{N \to \infty} \left[ \frac{1}{\sqrt{N+2} + \sqrt{N+1}} - \sqrt{2} + 1 \right] = 0 - \sqrt{2} + 1 = 1 - \sqrt{2} \]
**答案：** $\boxed{1 - \sqrt{2}}$
解析
步骤 1：重写通项
将通项重写为：\[ a_n = (\sqrt{n+2} - \sqrt{n+1}) - (\sqrt{n+1} - \sqrt{n}) \]
步骤 2：求前 $N$ 项和 $S_N$
求前 $N$ 项和 $S_N$：\[ S_N = \sum_{n=1}^{N} a_n = (\sqrt{N+2} - \sqrt{N+1}) - (\sqrt{2} - 1) \]
步骤 3：取极限
取极限：\[ \lim_{N \to \infty} S_N = \lim_{N \to \infty} \left[ \frac{1}{\sqrt{N+2} + \sqrt{N+1}} - \sqrt{2} + 1 \right] = 0 - \sqrt{2} + 1 = 1 - \sqrt{2} \]
将通项重写为：\[ a_n = (\sqrt{n+2} - \sqrt{n+1}) - (\sqrt{n+1} - \sqrt{n}) \]
步骤 2：求前 $N$ 项和 $S_N$
求前 $N$ 项和 $S_N$：\[ S_N = \sum_{n=1}^{N} a_n = (\sqrt{N+2} - \sqrt{N+1}) - (\sqrt{2} - 1) \]
步骤 3：取极限
取极限：\[ \lim_{N \to \infty} S_N = \lim_{N \to \infty} \left[ \frac{1}{\sqrt{N+2} + \sqrt{N+1}} - \sqrt{2} + 1 \right] = 0 - \sqrt{2} + 1 = 1 - \sqrt{2} \]